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B
cosπ8
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C
cosπ16
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D
cosπ32
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Solution
The correct option is Dcosπ32 Let cosx=√2+√2+√2+√22 ⇒4cos2x=2+√2+√2+√2⇒2cos2x=√2+√2+√2⇒4cos22x=2+√2+√2⇒2cos4x=√2+√2⇒4cos24x=2+√2⇒2cos8x=√2⇒4cos28x=2⇒2cos16x=0now⇒2cos24x=0⇒24x=π2∴x=π25 Hence cosπ32=√2+√2+√2+√22